Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-3x+6y &= 9 \\ -8x-6y &= -6\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-6y = 8x-6$ Divide both sides by $-6$ to isolate $y$ $y = {-\dfrac{4}{3}x + 1}$ Substitute this expression for $y$ in the first equation. $-3x+6({-\dfrac{4}{3}x + 1}) = 9$ $-3x - 8x + 6 = 9$ Simplify by combining terms, then solve for $x$ $-11x + 6 = 9$ $-11x = 3$ $x = -\dfrac{3}{11}$ Substitute $-\dfrac{3}{11}$ for $x$ back into the top equation. $-3( -\dfrac{3}{11})+6y = 9$ $\dfrac{9}{11}+6y = 9$ $6y = \dfrac{90}{11}$ $y = \dfrac{15}{11}$ The solution is $\enspace x = -\dfrac{3}{11}, \enspace y = \dfrac{15}{11}$.